Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $a \neq 0$. $z = \dfrac{10(4a + 3)}{7} \times \dfrac{-7}{16a^2 + 12a} $
Solution: When multiplying fractions, we multiply the numerators and the denominators. $z = \dfrac{ 10(4a + 3) \times -7 } { 7 \times (16a^2 + 12a) } $ $ z = \dfrac {-7 \times 10(4a + 3)} {7 \times 4a(4a + 3)} $ $ z = \dfrac{-70(4a + 3)}{28a(4a + 3)} $ We can cancel the $4a + 3$ so long as $4a + 3 \neq 0$ Therefore $a \neq -\dfrac{3}{4}$ $z = \dfrac{-70 \cancel{(4a + 3})}{28a \cancel{(4a + 3)}} = -\dfrac{70}{28a} = -\dfrac{5}{2a} $